KCET · Physics · Nuclear Physics
In a radioactive decay, an element \(\mathrm{z}^{\mathrm{X}}\) emits four \(\alpha\)-particles, three \(\beta\)-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are
- A \(\mathrm{Z}-8, \quad \mathrm{~A}-13\)
- B \(Z-11, A-16\)
- C \(\mathrm{Z}-5, \quad \mathrm{~A}-13\)
- D \(\mathrm{Z}-5, \quad \mathrm{~A}-16\)
Answer & Solution
Correct Answer
(D) \(\mathrm{Z}-5, \quad \mathrm{~A}-16\)
Step-by-step Solution
Detailed explanation
After \(\mathrm{n}_{\alpha} \quad \alpha\)-decays and \(\mathrm{n}_{\beta} \beta\)-decays, the respective atomic number and mass number of resulting nucleus are given by and \(\quad \mathrm{Z}^{\prime}=\mathrm{Z}-2 \mathrm{n}_{\alpha}+\mathrm{n}_{\beta}=\mathrm{Z}-8+3=\mathrm{Z}-5\) We have, \(\quad \mathrm{A}-4 \mathrm{n}_{\alpha}=\mathrm{A}-16\)
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